Question

Notify Me Bookmark

An electron moves through a uniform magnetic field. The magnetic field is given by B⃗⃗ = (0.00iˆ - 42.0jˆ + 62.0kˆ )mT. At a moment in time when the velocity is v = (0.00iˆ + vyĵ+ 87300.00kˆ) m/s, the magnetic force acting on the electron is F = (5.09iˆ + 0.00jˆ + 0.00kˆ) × 10-16 N. What is the value of vy? m/s

An electron moves through a uniform magnetic field. The magnetic field is given by B⃗⃗ = (0.00iˆ - 42.0jˆ + 62.0kˆ )mT. At a moment in time when the velocity is v = (0.00iˆ + vyĵ+ 87300.00kˆ) m/s, the magnetic force acting on the electron is F = (5.09iˆ + 0.00jˆ + 0.00kˆ) × 10-16 N. What is the value of vy? m/s

Image text
An electron moves through a uniform magnetic field. The magnetic field is given by B⃗⃗ = (0.00iˆ - 42.0jˆ + 62.0kˆ )mT. At a moment in time when the velocity is v = (0.00iˆ + vyĵ+ 87300.00kˆ) m/s, the magnetic force acting on the electron is F = (5.09iˆ + 0.00jˆ + 0.00kˆ) × 10-16 N. What is the value of vy? m/s

Detailed Answer

0 0

Please enter your email here. You will get email when this question is answered by our tutor.

Doubtrix Logo

Problem is not yet solved!

Click on Notify me to get email when question is answered.

Join Doubtrix with 7-days free trial

  • Icon Correct & Verified Answers
  • Icon No AI-Generated Answers
  • Icon Video Solution for Difficult Questions
  • Icon Work With Experts to Reach at Correct Answers
Notify me Login Sign Up

Similar/Related Questions

A proton moves through a uniform magnetic field. The magnetic field is given by B⃗ = (0.00iˆ - 29.4jˆ + 47.4kˆ )mT. At a moment in time when the velocity is v = (0.00iˆ + vyjˆ + 69600.00kˆ) m/s, the magnetic force acting on the proton is F = (5.42iˆ + 0.00jˆ + 0.00 kˆ) × 10-16 N. What is the value of vy?
Solution
0 0

An electron moves through a uniform magnetic field. The magnetic field is given by B→ = (0.00i^ − 90.5j^ + 50.9k^) mT. At a moment in time when the velocity is v→ = (0.00i^ + vyj^ + 98400.00k^) m/s, the magnetic force acting on the electron is F→ = (2.59i^ + 0.00j^ + 0.00k^)×10−16 N. What is the value of vy? m/s
Solution
0 0

A positively charged particle of mass 8.45×10−8 kg is traveling due east with a speed of 68.4 m/s and enters a 0.320−T uniform magnetic field. The particle moves through one-quarter of a circle in a time of 2.12×10−3 s, at which time it leaves the field heading due south. All during the motion the particle moves perpendicular to the magnetic field. (a) What is the magnitude of the magnetic force acting on the particle? (b) Determine the magnitude of its charge.
Solution
0 0